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I regard field-operators, whereby [itex]\Psi[/itex] is a fermionic-annihilation-operator and

[itex]\vec{A(\vec{r})}[/itex] is an electromagnetic-vector-potential.

Is it possible to do the following step?

$$

\nabla \vec{A} \Psi =\Psi \nabla \vec{A} + \vec{A}\nabla\Psi

$$

And if its correct, why?

Thx

Abby

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# Derivative of Field-Operator and Vector-Potential

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